\(2y^2-14y+24\)

\(=2y^2-6y-8y+24\)

\(=2y\left(y-3\right)-8\left(y-3\right)\)

\(=\left(y-3\right)\left(2y-8\right)\)

P(x)+Q(x)

=3x^2y-2x+5xy^2-7y^2+3xy^2-7y^2-9x^2y-x-5

=8xy^2-14y^2-6x^2y-3x-5

=>Chọn A

1/ \(x^2+x-90=\left(x^2-10x\right)+\left(9x-90\right)=x\left(x-10\right)+9\left(x-10\right)=\left(x-10\right)\left(x+9\right)\)

2/ \(2x^2+4xy+2y^2=\left(2x^2+2xy\right)+\left(2xy+2y^2\right)=2x\left(x+y\right)+2y\left(x+y\right)=\left(x+y\right)\left(2x+2y\right)\)

3/ \(2y^2-14y+24=2\left(y^2-7y+12\right)=2\left[\left(y^2-4y\right)+\left(12-3y\right)\right]=2\left[y\left(y-4\right)-3\left(y-4\right)\right]\)

\(=2\left(y-4\right)\left(y-3\right)\)

4/ \(x^8+x^4+1=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)

\(=\left(x^2+x+1\right)\left[\left(x^6-x^5+x^4\right)-\left(x^4-x^3+x^2\right)+\left(x^2-x+1\right)\right]\)

\(=\left(x^2+x+1\right)\left[x^4\left(x^2-x+1\right)\right]-x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\)

\(\Leftrightarrow\left(x-4\right)\left(x+\dfrac{6}{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{-6}{5}\end{matrix}\right.\)

Vậy pt có nghiệm là S=\(\left\{4;\dfrac{-6}{5}\right\}\)

=>5y^2-14y-24=0

=>5y^2-20y+6y-24=0

=>(y-4)(5y+6)=0

=>y=-6/5 hoặc y=4

Đưa một tỉ tao làm cho